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If \( \Large \tan \alpha = \left(1+2^{-x}\right)^{-1},\ \tan \beta = \left(1-2^{x+1}\right)^{-1},\ then\ \alpha - \beta \) equal

A) \( \Large \frac{ \pi }{6} \)
B) \( \Large \frac{ \pi }{4} \)
C) \( \Large \frac{ \pi }{3} \)
D) \( \Large \frac{ \pi }{2} \)

Correct Answer:


B) \( \Large \frac{ \pi }{4} \)

Description for Correct answer:
We have,

\( \Large \tan \left( \alpha + \beta \right) = \frac{\tan \alpha + \tan \beta }{1-\tan \alpha \tan \beta } \)

\( \Large \therefore \tan \alpha = \frac{1}{1+2^{-x}} \)

and \( \Large \tan \beta = \frac{1}{1+2^{x+1}} \)

\( \Large \frac{1}{1+\frac{1}{2^{x}}} + \frac{1}{1+2^{x+1}} \)

\( \Large \therefore \tan \left( \alpha + \beta \right) = \frac{2}{1+\frac{1}{2x}.\frac{1}{1+2^{x+1}}} \)

=> \( \Large \tan \left( \alpha + \beta \right) = \frac{2^{x}+2.2^{x+x}+2^{x}+1}{1+2^{x}+2.2^{x}+2.2^{x+x}-2^{x}} \)

=> \( \Large \tan \left( \alpha + \beta \right) = 1 \)

=> \( \Large \alpha + \beta = \frac{ \pi }{4} \)

Part of solved Trigonometric ratio questions and answers : >> Elementary Mathematics >> Trigonometric ratio

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